Homework #12 corrections

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1 Homework #2 corrections April 25, 205 (284) Let f() = 2 sin for 0 and f(0) = 0 (a) Use Theorems 283 and 284 to show f is differentiable at each a 0 and calculate f (a) Use, without proof, the fact that sin is differentiable and that cos is its derivative Proof Let g() = /, h() = sin and r() = 2 Then we can rewrite f() as f() = r()h(g()) We know that g is differentiable on all 0 (Eample 4 on page 227 in the tetbook), h is differentiable, and r is differentiable (Eample 3 on page 224 in the tetbook) So, by the Chain Rule, h g() = h(g()) is differentiable at for all 0, and we get ( ) (h g) () = h (g())g () = cos ( 2 ) We also have r () = 2 Thus, by the Product Rule, f() = r()h(g()) is differentiable for all 0, and, fiing = a, we obtain f (a) = r(a)h (g(a))g (a) + r (a)h(g(a)) = a 2 ( cos ( ) ( )) ( a a + 2a sin ) 2 a = cos ( ( a) + 2a sin ) a (b) Use the definition to show f is differentiable at = 0 and f (0) = 0 Proof By the definition of derivatives, we have f() f(0) 0 0 = 2 sin 0 = 0 sin The it above can be easily determined by the Squeeze Theorem In fact, since 0 sin(/), we have 0 sin ; ()

2 Math 320: Analysis I Spring 205 so if 0, the inequality () above implies, by the Squeeze Theorem, that 0 sin(/) = 0 So the it above eists and is finite; consequently, f is differentiable at = 0 and f (0) = 0 (c) Show f is not continuous at = 0 Proof By Eercise 284(a) above, for 0, ( f () = cos So we have to calculate the it [ ( ) ( )] cos + 2 sin 0 ) + 2 sin [ = cos 0 ( ) ( )] [ + 2 sin 0 ( )] But 0 cos(/) does not eist; therefore, 0 f () does not eist as well, and f is not continuous at = 0 (286) Let f() = sin for 0 and f(0) = 0 See Fig 93 (a) Observe f is continuous at = 0 by Eercise 79(c) Solution This has already been proven in class (b) Is f differentiable at = 0? Justify your answer Proof No, since the it does not eist (289) Let h() = ( 4 + 3) 7 (a) Calculate h () f() f(0) sin( = ) = sin Solution h () = 7( 4 + 3) 6 ( ) (b) Show how the Chain Rule justifies your computation in part (a) by writing h = g f for suitable f and g Solution Let f() = and g() = 7 ; then h = g f We have f () = and g () = 7 6 Thus, By the Chain Rule, h () = g (f())f () = [7( 4 + 3) 6 ]( ) (29) Determine whether the conclusion of the Mean Value Theorem holds for the following functions on the specified intervals If the conclusion holds, give an eample of a point satisfying () of Theorem 293 If the conclusion fails, state which hypotheses of the Mean Value Theorem fail 2

3 Math 320: Analysis I Spring 205 (a) 2 on [, 2] Solution Yes Let f() = 2 with dom(f) = [, 2] Then f () = 2 Furthermore, we have f( ) = and f(2) = 4, and so f(2) f( ) 2 ( ) = = Now we just have to let f () = 2 =, which implies = /2 (b) sin on [0, π] Solution Yes Let f() = sin, with dom(f) = [0, π] Then f () = cos Furthermore, we have f(0) = 0 = f(π), and so f(π) f(0) π 0 = 0 Now we let f () = cos() = 0, which implies = π/2 (c) on [, 2] Solution No Notice that But f () = f(2) f( ) 2 ( ) if < 0 0 if = 0 if > 0 = = 3, which is different than f () for every (, 2) The hypothesis that fails is the following: f() is not differentiable on 0 In effect, let f() =, with dom(f) = [, 2] Then But and thus f() f(0) 0 = = 0 =, f() f(0) 0 = = 0 + =, f() f(0) f() f(0) (d) on [, ] Solution No In fact, we have f () = / 2 ; however, f() f( ) ( ) = ( ) + =, and there is no (, ) such that f () = (f is always negative) The hypothesis that fails is this: f is discontinuous at = 0, since 0 = and 0 + = 3

4 Math 320: Analysis I Spring 205 (e) on [, 3] Solution Yes Let f() =, with dom(f) = [, 3] Then f () = Moreover, f() = and f(3) = 3 2, and hence f(3) f() 3 = /3 2 = 3 Now we put f () = 2 = 3, which results in = 3 (f) sgn() on [ 2, 2] Solution No Since sgn() = / for 0 and sgn(0) = 0, we have f () = 0 for 0, while f () is not defined for = 0 On the other hand, sgn(2) sgn( 2) 2 ( 2) = ( ) = 2 The hypothesis that fails is this: sgn is discontinuous at = 0, since 0 (292) Prove cos cos y y for all, y R 0 sgn() = and sgn() = + Proof First, let us begin with a trivial case: if = y, then cos cos y = 0 0 = = y, so clearly the inequality holds for this case In what follows, we will assume y Let f() = cos() Since f is differentiable on R, it is differentiable on any interval (, y) R By the Mean Value Theorem, there is a v (, y) such that f (v) = f() f(y) y We know that f () = sin ; so the equation above becomes sin v = cos cos y (2) y Taking the absolute value on both sides of equation () above, we get sin v = cos cos y cos cos y y = (3) y But sin for all R; this fact and equation (2) imply cos cos y y, or, equivalently, which is the desired result cos cos y y, 4

5 Math 320: Analysis I Spring 205 (297) (a) Suppose f is twice differentiable on an open interval I and f () = 0 for all I Show f has the form f() = a + b for suitable constants a and b Proof If f () = 0, then, by Corollary 294 in the tetbook, f is a constant function say, f () = a, where a I Let g be a function on I such that g() = a; then g is differentiable and g () = a = f () By Corollary 295 in the tetbook, this implies that f() = g()+b = a+b, for some constant b I This concludes the proof (b) Suppose f is three times differentiable on an open interval I and f = 0 on I What form does f have? Prove your claim Solution We claim that f() = a b + c, for constants a, b, c I In effect, if f () = 0, then, by Corollary 294 in the tetbook, f is a constant function defined by f () = a, for some a I Let g be a function on I such that g() = a; then g is differentiable and g () = a = f () By Corollary 295 in the tetbook, this implies that f () = g() + b = a + b, for some constant b I Finally, let h be a function on I defined by h() = a b; then h is differentiable on I and h () = a+b = f () By Corollary 295 in the tetbook, it follows that f() = h() + c = a b + c, for some constant c I Hence the claim is true (299) Show e e for all R Proof Let f() = e e; then f () = e e If >, f () > 0 and, by Corollary 297, f is strictly increasing; if <, f () < 0 and, again by Corollary 297, f is strictly decreasing; and if =, f () = 0 Since f is strictly decreasing for <, strictly increasing for >, and f is continuous on R, f() is a minimum for f But f() = e e = 0; therefore, f() = e e 0 for all R, which implies e e This concludes the proof (29) Show sin for all 0 Proof Let f() = sin ; then f () = cos Notice that for all 0, cos 0; therefore, f is increasing on [0, ) Since f(0) = 0 sin(0) = 0, it follows that f() = sin 0 for all 0 Hence sin for all 0, as we wanted to show Additional Problem: () Let a, b R Let f() = e a cos(b) and g() = e a sin(b) (i) Compute f () and g () Solution We have f () = be a sin(b) + ae a cos(b) and g () = be a cos(b) + ae a sin(b) 5

6 Math 320: Analysis I Spring 205 (ii) Use (i) to compute f and f Solution We have and f () = b 2 e a cos(b) abe a sin(b) abe a sin(b) + a 2 e a cos(b) = (a 2 b 2 )e a cos(b) 2abe a sin(b) f () = b(a 2 b 2 )e a sin(b) + a(a 2 b 2 )e a cos(b) 2ab 2 e a cos(b) 2a 2 be a sin(b) = (a 2 b 2 )e a (a cos(b) b sin(b)) 2abe a (b cos(b) a sin(b)) 6